§ME 231 Homework 9

Sep 25, 2025

§6.28

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3, is stirred until its temperature is 500 K. Assuming the ideal gas model for the air, and ignoring kinetic and potential energy, determine (a) the final pressure, in bar, (b) the work, in kJ, and (c) the amount of entropy produced, in kJ/K. Solve usinga. data from Table A-22.b. constant cv read from Table A-20 at 400 K.
W˙=0 \dot{W}=0
Q˙=0 \dot{Q}=0
T1=300K T_1=300K
p1=2bar p_1=2bar
V=2m3 V=2m^3
T2=500K T_2=500K

§a.

The pressure fo the second state can be fetched using the ideal gas law:
pV=mRT pV=mRT
pVT=mR \frac{pV}{T}=mR
pT=mRV=const=p1T1=p2T2 \frac{p}{T}=\frac{mR}{V}=\text{const}=\frac{p_1}{T_1}=\frac{p_2}{T_2}
p2=p1T2T1=2bar∗500K300K=3.333bar p_2=\frac{p_1 T_2}{T_1}=\frac{2bar * 500K}{300K}=\boxed{3.333bar}
Q−W=ΔE=ΔU Q-W=\Delta E=\Delta U
−W=ΔU=U2−U1=m(u2−u1) -W=\Delta U=U_2-U_1=m(u_2-u_1)
From table A-22 where T=300KT=300K
u1=214.07kJkg u_1=214.07\frac{kJ}{kg}
u2u_2 must be found using interpolation. Excerpt from table A-22:
TT uu
440K440K 315.30kJkg315.30\frac{kJ}{kg}
750K750K 551.99kJkg551.99\frac{kJ}{kg}
u2=315.30+(551.99−315.30)500−440750−440=361.11kJkg u_2=315.30+(551.99-315.30)\frac{500-440}{750-440}=361.11\frac{kJ}{kg}
From table A-1 for air:
M=28.97kgkmol M=28.97\frac{kg}{kmol}
Rˉ=8.3145Jmol∗K \bar{R}=8.3145\frac{J}{mol*K}
R=RˉM=8.3145Jmol∗K28.97kgkmol=287Jkg∗K R=\frac{\bar{R}}{M}=\frac{8.3145\frac{J}{mol*K}}{28.97\frac{kg}{kmol}}=287\frac{J}{kg*K}
pV=mRT pV=mRT
p1V=mRT1 p_1V=mRT_1
m=p1VRT1=2bar∗2m3287Jkg∗K∗300K=4.646kg m=\frac{p_1V}{RT_1}=\frac{2bar*2m^3}{287\frac{J}{kg*K} * 300K}=4.646kg
W=−m(u2−u1)=−4.646kg∗(361.11kJkg−214.07kJkg)=−683.1kJ W=-m(u_2-u_1)=-4.646kg * (361.11\frac{kJ}{kg} - 214.07\frac{kJ}{kg})=\boxed{-683.1kJ}
A negative answer for WW makes sense because work was done to the system.From table A-22:
s°(T1)=s°(300K)=1.70203kJkg∗K s\degree(T_1)=s\degree(300K)=1.70203\frac{kJ}{kg*K}
s°(T2)s\degree(T_2) must be found using interpolation. Excerpt from table A-22:
TT s°(T)s\degree(T)
440K440K 2.08870kJkg∗K2.08870\frac{kJ}{kg*K}
750K750K 2.64737kJkg∗K2.64737\frac{kJ}{kg*K}
s°(T2)=s°(500K)=2.08870+(2.64737−2.08870)500−440750−440=2.19683kJkg∗K s\degree(T_2)=s\degree(500K)=2.08870+(2.64737-2.08870)\frac{500-440}{750-440}=2.19683\frac{kJ}{kg*K}
ΔS=∂QT+Δσ \Delta S = \frac{\partial Q}{T} + \Delta \sigma
Δσ=ΔS=S2−S1=S°(T2)−S°(T1)−Rln⁡p2p1 \Delta \sigma = \Delta S = S_2-S_1=S\degree(T_2)-S\degree(T_1)-R \ln \frac{p_2}{p_1}
Δσ=m(s°(T2)−s°(T1)−Rln⁡p2p1) \Delta \sigma=m(s\degree(T_2)-s\degree(T_1)-R \ln \frac{p_2}{p_1})
Δσ=4.646kg∗(2.19683kJkg∗K−1.70203kJkg∗K−287Jkg∗K∗ln⁡3.333bar2bar)=1.618kJ/K \Delta \sigma=4.646kg * \left(2.19683\frac{kJ}{kg*K}-1.70203\frac{kJ}{kg*K}-287\frac{J}{kg*K} * \ln \frac{3.333bar}{2bar}\right)=\boxed{1.618kJ/K}

§b.

From table A-20 where T=400KT=400K for air:
cv=0.726kJkg∗K c_v=0.726\frac{kJ}{kg*K}
Regardless of a constant cvc_v, I would use the ideal gas law equation the same way to find p2p_2 (thus mm too) as in part (a).
p2=3.333bar p_2=\boxed{3.333bar}
m=4.646kg m=4.646kg
Q−W=ΔE=ΔU Q-W=\Delta E=\Delta U
−W=ΔU=mcv(T2−T1) -W=\Delta U=m c_v (T_2 - T_1)
W=ΔU=−mcv(T2−T1)=−4.646kg∗0.726kJkg∗K(500K−300K)=−674.6kJ W=\Delta U=- m c_v (T_2 - T_1)=- 4.646kg * 0.726\frac{kJ}{kg*K} (500K - 300K)=\boxed{-674.6kJ}
That's pretty close!
σ=S2−S1=m(s2−s1)=m(cvln⁡T2T1+Rln⁡VV)=mcvln⁡T2T1=4.646kg∗0.726kJkg∗Kln⁡500K300K \sigma=S_2-S_1=m(s_2-s_1)=m(c_v \ln \frac{T_2}{T_1} + R \ln \frac{V}{V})=m c_v \ln \frac{T_2}{T_1}=4.646kg * 0.726\frac{kJ}{kg*K} \ln \frac{500K}{300K}
σ=1.723kJ/K \sigma=\boxed{1.723kJ/K}
Also pretty dang close! But I would trust part (a) more lol.

§6.32

Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per kg of steam for the process? Calculate the amount of entropy produced, in kJ/K per kg of steam. What is the maximum theoretical work that could be obtained from the given initial state to the same final pressure? Show both processes on a properly labeled sketch of the T–s diagram.
Q=0 Q=0
p1=100bar p_1=100bar
T1=360°C T_1=360\degree C
p2=1bar p_2=1bar
T2=160°C T_2=160\degree C
From table A-4 at p=100barp=100bar and T=360°CT=360\degree C:
u1=2729.1kJkg u_1=2729.1\frac{kJ}{kg}
From table A-4 at p=1barp=1bar and T=160°CT=160\degree C:
u2=2597.8kJkg u_2=2597.8\frac{kJ}{kg}
q−w=Δe=Δu q-w=\Delta e=\Delta u
−w=u2−u1 -w=u_2-u_1
w=u1−u2=2729.1kJkg−2597.8kJkg=131.3kJkg w=u_1-u_2=2729.1\frac{kJ}{kg}-2597.8\frac{kJ}{kg}=\boxed{131.3\frac{kJ}{kg}}
From table A-4 at p=100barp=100bar and T=360°CT=360\degree C:
s1=6.0060kJkg∗K s_1=6.0060\frac{kJ}{kg*K}
From table A-4 at p=1barp=1bar and T=160°CT=160\degree C:
s2=7.6597kJkg∗K s_2=7.6597\frac{kJ}{kg*K}
ΔS=∂QT+Δσ \Delta S = \frac{\partial Q}{T} + \Delta \sigma
Δσ=ΔS \Delta \sigma = \Delta S
Δσm=Δs=s2−s1=7.6597kJkg∗K−6.0060kJkg∗K=1.6537kJkg∗K \frac{\Delta \sigma}{m} = \Delta s = s_2 - s_1=7.6597\frac{kJ}{kg*K}-6.0060\frac{kJ}{kg*K}=\boxed{1.6537\frac{kJ}{kg*K}}
Ideally in an isentropic process, the −W-W would be the change in enthalpy.From table A-4 at p=100barp=100bar and T=360°CT=360\degree C:
h1=2962.1kJkg h_1=2962.1\frac{kJ}{kg}
From table A-4 at p=1barp=1bar and T=160°CT=160\degree C:
h2=3195.9kJkg h_2=3195.9\frac{kJ}{kg}
w=h1−h2=2962.1kJkg−3195.9kJkg=−233.8kJkg w=h_1-h_2=2962.1\frac{kJ}{kg}-3195.9\frac{kJ}{kg}=-233.8\frac{kJ}{kg}

§6.60

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occurs at an average outer surface temperature of 315 K at the rate of 30 kJ per kg of air flowing. Kinetic and potential energy effects are negligible. Assuming the air is modeled as an ideal gas with variations in specific heat, determine (a) the rate power is developed, in kJ per kg of air flowing, and (b) the rate of entropy production within the turbine, in kJ/K per kg of air flowing.
p1=400kPa p_1=400kPa
T1=980K T_1=980K
m˙out=m˙in=m˙ \dot{m}_{out}=\dot{m}_{in}=\dot{m}
p2=100kPa p_2=100kPa
T2=670K T_2=670K
Tout=315K T_{out}=315K
q=30kJ/kg q=30kJ/kg
Q˙−W˙+m˙(h1+12v12+gz1)−m˙(h2+12v22+gz2)=0 \dot{Q}-\dot{W}+\dot{m}(h_1+\frac{1}{2}v_1^2+gz_1)-\dot{m}(h_2+\frac{1}{2}v_2^2+gz_2)=0
q−w+h1+12v12+gz1−h2−12v22−gz2=0 q-w+h_1+\frac{1}{2}v_1^2+gz_1-h_2-\frac{1}{2}v_2^2-gz_2=0
q−w+h1−h2=0 q-w+h_1-h_2=0
−w=−q−h1+h2 -w=-q-h_1+h_2
w=q+h1−h2 w=q+h_1-h_2
From table A-22 at T=980KT=980K:
h1=1023.25kJ/kg h_1=1023.25kJ/kg
s°1=2.94468kJkg∗K s\degree_1=2.94468\frac{kJ}{kg*K}
Excerpt from table A-22 at T=670KT=670K:
TT hh s°s\degree
440K440K 441.61kJ/kg441.61kJ/kg 2.08870kJkg∗K2.08870\frac{kJ}{kg*K}
750K750K 767.29kJ/kg767.29kJ/kg 2.64737kJkg∗K2.64737\frac{kJ}{kg*K}
h2=441.61+(767.29−441.61)670−440750−440=683.24kJ/kg h_2=441.61+(767.29-441.61)\frac{670-440}{750-440}=683.24kJ/kg
s°2=2.08870+(2.64737−2.08870)670−440750−440=2.50320kJkg∗K s\degree_2=2.08870+(2.64737-2.08870)\frac{670-440}{750-440}=2.50320\frac{kJ}{kg*K}
w=30kJ/kg+1023.25kJ/kg−683.24kJ/kg=370kJ/kg w=30kJ/kg+1023.25kJ/kg-683.24kJ/kg=\boxed{370kJ/kg}
s2−s1=∫∂qT+σ≈qTout+σ s_2-s_1=\int \frac{\partial q}{T} + \sigma \approx \frac{q}{T_{out}} + \sigma
σ=s2−s1−qTout=s°2−s°1−Rln⁡p2p1−qTout \sigma = s_2 - s_1 - \frac{q}{T_{out}} = s\degree_2 - s\degree_1 - R \ln \frac{p_2}{p_1} - \frac{q}{T_{out}}
R=287Jkg∗K R=287\frac{J}{kg*K}
σ=2.50320kJkg∗K−2.94468kJkg∗K−287Jkg∗K∗ln⁡100kPa400kPa−30kJ/kg315K=0.086kJkg∗K \sigma = 2.50320\frac{kJ}{kg*K} - 2.94468\frac{kJ}{kg*K} - 287\frac{J}{kg*K} * \ln \frac{100kPa}{400kPa} - \frac{30kJ/kg}{315K}=\boxed{0.086\frac{kJ}{kg*K}}

§6.62

Air at 200 kPa, 52°C, and a velocity of 355 m/s enters an insulated duct of varying cross-sectional area. The air exits at 100 kPa, 82°C. At the inlet, the cross-sectional area is 6.57 cm2. Assuming the ideal gas model for the air, determinea. the exit velocity, in m/s.b. the rate of entropy production within the duct, in kW/K.
p1=200kPa p_1=200kPa
T1=52°C=325.2K T_1=52\degree C=325.2K
v1=355m/s v_1=355m/s
Q=W=0 Q=W=0
p2=100kPa p_2=100kPa
T2=82°C=355.2K T_2=82\degree C=355.2K
A1=6.57cm2 A_1=6.57cm^2
From table A-22 at T=325.2K≈325KT=325.2K \approx 325K:
h1=325.31kJ/kg h_1=325.31kJ/kg
s°1=1.78249kJkg∗K s\degree_1=1.78249\frac{kJ}{kg*K}
From table A-22 at T=355.2K≈360KT=355.2K \approx 360K:
h2=360.58kJ/kg h_2=360.58kJ/kg
s°2=1.88543kJkg∗K s\degree_2=1.88543\frac{kJ}{kg*K}
Q˙−W˙+m˙(h1+12v12+gz1)−m˙(h2+12v22+gz2)=0 \dot{Q}-\dot{W}+\dot{m}(h_1+\frac{1}{2}v_1^2+gz_1)-\dot{m}(h_2+\frac{1}{2}v_2^2+gz_2)=0
m˙(h1+12v12)−m˙(h2+12v22)=0 \dot{m}(h_1+\frac{1}{2}v_1^2)-\dot{m}(h_2+\frac{1}{2}v_2^2)=0
h1+12v12−h2−12v22=0 h_1+\frac{1}{2}v_1^2-h_2-\frac{1}{2}v_2^2=0
12v22=h1+12v12−h2 \frac{1}{2}v_2^2=h_1+\frac{1}{2}v_1^2-h_2
v22=2h1+v12−2h2 v_2^2=2h_1+v_1^2-2h_2
v2=2h1+v12−2h2 v_2=\sqrt{2h_1+v_1^2-2h_2}
v2=2∗325.31kJ/kg+(355m/s)2−2∗360.58kJ/kg=235.6m/s v_2=\sqrt{2 * 325.31kJ/kg + (355m/s)^2 - 2 * 360.58kJ/kg}=\boxed{235.6m/s}
s2−s1=∫∂qT+σ=σ s_2-s_1=\cancel{\int \frac{\partial q}{T}} + \sigma = \sigma
R=287Jkg∗K R=287\frac{J}{kg*K}
σ=s2−s1=s°2−s°1−Rln⁡p2p1 \sigma = s_2 - s_1 = s\degree_2 - s\degree_1 - R \ln \frac{p_2}{p_1}
σ=1.88543kJkg∗K−1.78249kJkg∗K−287Jkg∗K∗ln⁡100kPa200kPa=0.3019kJkg∗K \sigma = 1.88543\frac{kJ}{kg*K} - 1.78249\frac{kJ}{kg*K} - 287\frac{J}{kg*K} * \ln \frac{100kPa}{200kPa}=\boxed{0.3019\frac{kJ}{kg*K}}